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大神求解这道题!!!!谢谢!!!
1个回答
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(1)1=∫[-∞,+∞]f(x)dx=∫[0,1]xdx+∫[1,2](a-x)dx
=x²/2 |[0,1]+(ax-x²/2)|[1,2]
=1/2 -0+(2a-2)-(a-½)
=a-1
故a=2
(2)E(X)=∫[-∞,+∞]xf(x)dx=∫[0,1]x²dx+∫[1,2](2x-x²)dx
=x^3 /3|[0,1]+(x²-x^3/3)|[1,2]
=1/3-0+(4-8/3)-(1-1/3)
=1
=x²/2 |[0,1]+(ax-x²/2)|[1,2]
=1/2 -0+(2a-2)-(a-½)
=a-1
故a=2
(2)E(X)=∫[-∞,+∞]xf(x)dx=∫[0,1]x²dx+∫[1,2](2x-x²)dx
=x^3 /3|[0,1]+(x²-x^3/3)|[1,2]
=1/3-0+(4-8/3)-(1-1/3)
=1
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谢谢!!
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